Problem Set 3 Solutions

Transcription

1 Chemistry 36 Dr Jean M Standard Problem Set 3 Solutions 1 Verify for the particle in a one-dimensional box by explicit integration that the wavefunction ψ x) = π x ' sin ) is normalized To verify that ψ x) is normalized we have to show that ψ x) dx = 1 Note that the integration range is from x= to x= since the particle in a box wavefunctions vanish outside that range Substituting the integral becomes ψ x) dx = = π x sin' * ) sin π x ' * dx ) π x sin' * dx ) The integral that we need can be found for example in the CRC or in most any table of integrals: sin bx dx = x sinbx 4b Setting b = π and substituting the normalization integral becomes ψ x) dx = sin π x ' * dx ) / = x 1 4 ' π sin 4π x 1 ' * 1 ) 1 * - ) 1

2 1 continued Evaluating the expression at the limits ψ x) dx = = = ψ x) dx = 1 Therefore the wavefunction ψ x) is normalized / x 1 4 π sin 4π x ) ) ' * 1 ' + - * / sin 4π ) - 8π 1 8π sin / 5 3 ) / 3 1 [ ]

3 3 Calculate the energy level spacings in Joules between the ground n=1) and first excited n=) levels for the following cases a) An electron confined to a one-dimensional box of width 5 Å The energy level spacing ΔE between the ground and first excited state is ΔE = E E 1 Using the particle in a box energy expression we can obtain an equation for the energy level spacing Substituting ΔE = E E 1 = h 8m 1 h 8m ΔE = ΔE = E E 1 = 3h 8m Js) ) m) kg ΔE = E E 1 = J b) A baseball with a mass of 14 g confined to a one-dimensional box of width 1 meter Using the same expression from part a) for the energy level spacing we have Substituting ΔE = E E 1 = h 8m 1 h 8m ΔE = 3h 8m ) ΔE = E E 1 = Js ) 1 m) 8 1 kg ΔE = E E 1 = J Notice how much smaller the energy level spacing is for the macroscopic object Energy level spacings this small are not measurable and therefore the energy level spacings are so small as to be effectively continuous for macroscopic objects

4 4 3 An electron in a box of width undergoes a transition from the lowest energy level n=1) to the first excited level n=) The wavelength of light absorbed in this transition was determined to be 65 nm Calculate the width of the box For a transition from n=1 to n= the energy difference ΔE is ΔE = E E 1 A photon with an energy corresponding to ΔE would have a frequency given by E photon = ΔE = hν Since for light λν = c we can substitute ν = λ and obtain an expression for the energy difference c ΔE = E E 1 = hc λ Substituting ΔE = E E 1 = ΔE = E E 1 = J Js) ms 1 68 nm) 1-9 m ' * 1 nm ) ) Then we can use the particle in a box energies to obtain an expression for the energy difference ΔE = E E 1 = h 8m 1 h 8m ΔE = Solving for the width of the box yields 3h 8m = # 3h 8mΔE ' 1/ Substituting = = # 3h 8mΔE ' 1/ Js) ) J) kg - = m or 7866 Å / 1/

5 5 4 Determine the most probable location for the particle in the ground state of the one-dimensional particle in a box Also determine the most probable location for the first excited state of the particle in a box To obtain the most probable value we must look at the probability density ψ x) For the ground state of the particle in a box the probability density is given by ) = π x sin ψ 1 x A plot of this probability density is shown below ' ) 15 ψ 1 x) x/ Note that the maximum of this function occurs at x = ; this is the most probable value In this case the maximum can be found by inspection For more general situations to get the maximum value we would have to take the derivative set it equal to zero and solve For the first excited state of the particle in a box the probability density is given by A plot of this probability density is given below ψ x) = π x ' sin ) 15 ψ x) x/ For this function we see from the plot that there are two maxima each with identical amplitude By inspection these maxima occur at x = 4 and x = 3 4 ; these two values are therefore the most probable values

6 6 5 For a particle in a one-dimensional box of width determine the probability of finding the particle in the right third of the box between /3 and ) if the particle is in the ground state Since the probability is given by ψ x) dx if we want the total probability of finding the particle between /3 and we must add up the probability for all the points from /3 to Since x is a continuous variable the sum is really an integral from /3 to Probability = / 3 ψ x) dx Substituting the particle in a box wavefunction for the ground state ψ 1 x) = integral becomes sin π x ' ) the probability Probability = sin # π x dx ' / 3 This integral can be evaluated using tables From the CRC Handbook or any other table of integrals we find the indefinite integral: sin bx dx = x sin bx 4b Replacing b with π yields * Probability = x + sin π x '- )/ 4 π / ' / ) / / 3 Finally evaluating the expression at the limits leads to Probability = π sin # 4π 3 ' Evaluating the numerical value of the sine function in the expression above the probability is Probability = = 1955

7 7 6 The 135-hexatriene molecule is a conjugated molecule with 6 pi electrons Consider the pi electrons free to move back and forth along the molecule through the delocalized pi system Using the particle in a box approximation treat the carbon chain as a linear one-dimensional "box" Allow each energy level in the box to hold pi electrons Treating 135-hexatriene as a linear chain we would have the structure The pi electrons are free to move through the conjugated system; hence the length of the molecule would represent the "box" We can add up the lengths of the single and double bonds in the molecule to give the width of the box For 135-hexatriene this gives where r represents the bond length C 1 C C 3 C 4 C 5 C 6 = 3r C =C + r C C The energy of the pi electrons in the molecule would be represented as E n = n h n = 1 3 8m where is the box width defined above and m is the electron mass a) Calculate the energy of the highest filled level using 154 Å as the carbon-carbon single bond length and 135 Å as the carbon-carbon double bond length Using r C =C = 135 Å and r C C = 154 Å the width of the box can be calculated from the equation above = 3135 Å) Å) = 713 Å or m) Using the formula above with = m for the energy leads to the energy diagram shown below n=5 n=4 n=3 n= n=1

8 8 6 a) continued 135-hexatriene has 6 pi electrons Placing two electrons in each energy level fills the levels up to the n=3 level as shown in the figure above So n=3 is the highest filled level Computing the energy of the highest filled level E 3 = Js) ) m) ) kg = J E 3 = J b) Determine the energy of the lowest unfilled level The lowest unfilled level corresponds to n=4 Calculating the energy of this state gives E 4 = 4 h 8ma ) E 4 = J E 4 = J c) Calculate the wavelength for an electronic transition from the highest filled level to the lowest unfilled level using your answers from parts a) and b) Compare your result to the experimental ultraviolet absorption maximum of 68 nm For an electronic transition from n=3 to n=4 the energy difference ΔE is ΔE = E 4 E 3 A photon with an energy corresponding to ΔE would have a frequency given by E photon = ΔE = hν Since for light λν = c we can substitute ν = λ and solve for the wavelength to give c λ = hc ΔE = hc E 4 E 3 Inserting numerical values leads to λ = Js) ms 1 ) J J) λ = m or λ = 395nm This result is fairly close to the experimental wavelength λ exp = 68 nm with about an 11 error This is surprisingly good agreement given the crudeness of the model

9 9 7 Verify by explicit integration that ψ and ψ 3 for the particle in a one-dimensional box are orthogonal For a particle in a 1D box the wavefunctions are given by ψ n x) = nπx ' sin ) To show that ψ and ψ 3 are orthogonal the integral of the product of the two functions including the complex conjugate of one of the functions if they are not real functions) must be zero Evaluating the integral in this case both functions are real) we have ψ * x)ψ 3 x) dx = ψ x)ψ 3 x) dx = sin πx ' * sin 3πx ' * dx ) ) Note that the integration range is from x= to x= since the particle in a box wavefunctions vanish outside that range The indefinite integral that we need can be found for example in the CRC or in most any table of integrals: sin mx sin nx dx = sin m n)x m n ) sin m + n)x m + n ) Setting m = π and n = 3π we have m n = π and m + n = 5π Substituting ψ x)ψ 3 x) dx = πx ) sin + - π ' * 1π 5πx )/ sin + 1 ' * Using the identity sin x) = sin x and evaluating the integral at the limits yields ψ x)ψ 3 x) dx = ' π sin π 1π ) sin 5π * + ' π sin 1π sin ) * + Since sin π = sin 5π = sin = the integral simplifies to ψ x)ψ 3 x) dx = Therefore the functions ψ x) and ψ 3 x) are orthogonal

10 1 8 Determine the average value of the position x for the ground state of the one-dimensional particle in a box Compare your result with the most probable location The average value of the position is given by x = ψ * 1 x) x ˆ ψ x) dx = ψ * 1 1 x) x ψ 1 x) dx where the definition of the position operator ˆ x = x has been used In addition the limits of the integral are x= to x= because the wavefunction vanishes outside this range The ground state wavefunction for the particle in a box is given by ψ 1 x) = sin π x ' ) Upon substitution the expression for the average value of the position becomes x = = = ψ 1 * x) x ψ 1 x) dx sin π x ' * x ) sin π x ' * dx ) x sin π x ' * dx ) From the CRC Handbook or handout of integrals) x sin α x dx = x 4 x sin α x 4α cos α x 8α Replacing α by π the average value becomes x = * + x 4 x sin π x ' ) 4 π ' ) cos π x '- )/ / 8 π ' / ) /

11 11 8 continued Evaluating the expression at the limits yields the average value of the position for the ground state of the particle in a box: x = = 4 sin π 4π ) cos π ) ' ) 8π sin 4 ' 8π ) ' 8π ) ) 8π cos ) ' ) x = For the ground state of the particle in a box the average value and most probable value of x are identical the most probable value of / was found in problem 4) This will not always be the case as we will see in next problem

12 1 9 Repeat problem 8 for the first excited state of the particle in a box Does your result agree this time with the most probable location for this state? For the first excited state of the particle in a box the wavefunction is ψ x) = π x ' sin ) Thus the average value of x for this state is x = = x = ψ * x) x ψ x) dx π x sin' * x ) x sin π x ' * dx ) π x sin' * dx ) We can use the same integral as we did in the previous problem; in this case we replace α by π : x = * + x 4 x sin 4π x ' ) 4 π ' ) cos 4π x '- )/ / 8 π ' / ) / Evaluating this expression at the limits gives the average value of the position for the first excited state x = = 4 sin 4π ) 8π 3π cos 4π ) ' ) sin 4 ' 3π ) ' 3π ) ) 3π cos ) ' ) x = This is exactly the same result that we got for the ground state This will not always happen here we got the same result because of the symmetry of the potential energy and as a result the symmetry of the wavefunction about x=/ From problem 4 we found that the most probable values for the first excited state occur at x = 4 and x = 3 4 Note that for the first excited state of the particle in a box the average value and most probable values of x are not the same

13 1 Determine the average value of the momentum p x for the ground state of the one-dimensional particle in a box 13 The average value of the momentum is given by p x = ψ * 1 x) p ˆ x ψ 1 x) dx = i! ψ * d 1 x) dx ψ 1x) dx where the definition of the momentum operator p ˆ x = i! d has been used In addition the limits of the dx integral again are x= to x= because the wavefunction vanishes outside this range The ground state wavefunction for the particle in a box is given by ψ 1 x) = sin π x ' ) Upon substitution the expression for the average value of the momentum becomes p x = i! ψ 1 * x) = i! = i! d dx ψ 1x) dx sin π x ) + d - ' * dx / sin π x ) + 1 dx ' * sin π x ) + d - ' * dx sin π x ) + 1 / ' * dx Next the derivative must be evaluated d dx ) sin # π x * - + ' = π cos # π x ' Substituting the average value of the momentum is p x = i!π sin π x ' ) cos π x ' ) dx From the CRC Handbook or handout of integrals) sin bx cosbx dx = sin bx b

14 14 1 continued Replacing b by π the average value of momentum becomes p x = i!π sin π x π Evaluating this expression at the limits gives the average value of momentum for the ground state p x = i!π = i!π [ ] + p x = π sin π ) + i!π π sin ) i!π [ ] For the ground state of the particle in a box the average value of momentum is This is actually true for any state of the particle in a box and reflects the idea that motion in the positive x-direction corresponding to positive values of momentum and motion in the negative x-direction corresponding to negative values of momentum are equally probable Thus the positive and negative values of momentum cancel yielding an average value of